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d and f Block Elements - Lanthanoids and Actinoids

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The ff-block elements are called inner transition elements because they involve the filling of the (n2)f(n-2)f subshell. They consist of two series: Lanthanoids (4f4f series) and Actinoids (5f5f series).

General electronic configuration of Lanthanoids is [Xe]4f1145d016s2[Xe] 4f^{1-14} 5d^{0-1} 6s^2 and for Actinoids it is [Rn]5f1146d017s2[Rn] 5f^{1-14} 6d^{0-1} 7s^2.

Lanthanoid Contraction: The steady decrease in the atomic and ionic radii (specifically M3+M^{3+} ions) with increase in atomic number. This is caused by the poor shielding effect of 4f4f electrons against the increasing nuclear charge.

Consequences of Lanthanoid Contraction: It leads to the similarity in the chemical properties of elements of the second (4d4d) and third (5d5d) transition series, such as ZrZr and HfHf having nearly identical radii.

Oxidation States: The most common oxidation state for both series is +3+3. However, Lanthanoids exhibit +2+2 and +4+4 if they lead to stable f0,f7f^0, f^7, or f14f^{14} configurations (e.g., Ce4+Ce^{4+} is f0f^0 and Eu2+Eu^{2+} is f7f^7).

Actinoids show a wider range of oxidation states (up to +7+7) compared to Lanthanoids because the energy difference between 5f,6d5f, 6d, and 7s7s subshells is very small, allowing more electrons to participate in bond formation.

Actinoid Contraction: Similar to Lanthanoid contraction, but more pronounced due to even poorer shielding by 5f5f electrons compared to 4f4f electrons.

Chemical Reactivity: Lanthanoids are highly electropositive and resemble Calcium in reactivity. Actinoids are even more reactive, especially when finely divided, and many are radioactive (transuranic elements).

📐Formulae

[Xe]4f1145d016s2[Xe] 4f^{1-14} 5d^{0-1} 6s^2

[Rn]5f1146d017s2[Rn] 5f^{1-14} 6d^{0-1} 7s^2

μ=n(n+2) B.M.\mu = \sqrt{n(n+2)} \text{ B.M.}

Basicity order: La(OH)3>Ce(OH)3>>Lu(OH)3\text{Basicity order: } La(OH)_3 > Ce(OH)_3 > \dots > Lu(OH)_3

💡Examples

Problem 1:

Explain why Ce4+Ce^{4+} is a strong oxidizing agent even though it has a stable 4f04f^0 configuration.

Solution:

Ce4+Ce^{4+} (4f04f^0) tends to revert to the most stable oxidation state of the Lanthanoid series, which is +3+3 (4f14f^1).

Explanation:

Since Ce4+Ce^{4+} gains an electron to become Ce3+Ce^{3+}, it acts as a strong oxidizing agent. The EE^{\circ} value for Ce4+/Ce3+Ce^{4+}/Ce^{3+} is +1.74 V+1.74 \text{ V}.

Problem 2:

Compare the basicity of La(OH)3La(OH)_3 and Lu(OH)3Lu(OH)_3.

Solution:

La(OH)3La(OH)_3 is more basic than Lu(OH)3Lu(OH)_3.

Explanation:

Due to Lanthanoid contraction, the size of M3+M^{3+} ions decreases from La3+La^{3+} to Lu3+Lu^{3+}. According to Fajan's rules, a decrease in size increases the covalent character of the MOHM-OH bond. Therefore, Lu(OH)3Lu(OH)_3 is more covalent and less basic than La(OH)3La(OH)_3.

Problem 3:

Calculate the magnetic moment of La3+La^{3+} (Z=57Z=57).

Solution:

μ=0 B.M.\mu = 0 \text{ B.M.}

Explanation:

The electronic configuration of LaLa is [Xe]5d16s2[Xe] 5d^1 6s^2. For La3+La^{3+}, the configuration is [Xe]4f0[Xe] 4f^0, meaning there are n=0n=0 unpaired electrons. Using μ=n(n+2)\mu = \sqrt{n(n+2)}, we get 00.

Lanthanoids and Actinoids - Revision Notes & Key Formulas | ICSE Class 12 Chemistry